Originally posted by Aetherael
i'll use p and q as the generating numbers (p>q), and let [b]a = 2pq, b = p^2-q^2, and c = p^2 + q^2
a perfect square is always either 0, 1 or 4 (mod5). it also follows that if x^2 = 0 (mod 5), then x=0 (mod 5).
now let's look at the generating formulas for pythagorean triples: if p=0 (mod5) or q=0 (mod5) then clearly 2pq = 0 (mod 5) and so g of p^2-q^2 as (p-q)(p+q)... i'm too lazy to try and figure it out though 🙂[/b]
The conjecture of the author of this thread that all triplets arise from the basic triplet (3,4,5) is indeed true.
Given any primitive (independent pythagorean triplet in its lowest multiple form) triplet (a,b,c) , three new primitive triples can be generated as follows:-
(a1,b1,c1) = (a,b,c)U......................................(1)
(a2,b2,c2) = (a,b,c)A.....................................(2)
(a3,b3,c3) = (a,b,c)D.......................................(3)
Where U A D are defined as the following three matrices respectively : -
U = 1,2,2; ....A =-1,-2,-2;....D = 1,2,2
......2,1,2.......... 2, 1, 2 ..........-2,-1,-2
......2,2,3;.........2, 2, 3;..........2,2,3;
It has been proven that any Pythagorean triplet (p,q,r) can be a primitive Pythagorean triplet iff (p,q,r) = (3,4,5)M;
where M is a finite product of the matrices U , A and D.
In short M can be any matrix from the following pyramidical array of matrices:-
U, .. A, .. D,
UU,UA,UD, .. AU, AA, AD, .. DU, DA, DD,
UUU,UUA,UUD,.. UAU,UAA,UAD,..UDU,UDA,UDD ... AUU,AUA,AUD,etc.
The n-th row in this pyramydical array has 3^n matrices.
Thus all primitive (independent and distinct i.e. in their lowest multiple form) Pythagorean triplets do indeed arise from the basic triplet (3,4,5). You can see the link given below.
http://mathworld.wolfram.com/PythagoreanTriple.html